TikTok Machine Learning Engineer Interview Questions

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Rajesh V. - "As a PM i received a feedback from my program manager on my style of verbal communication. It is about me speaking faster when i wanted to get away with a topic that i wasn't confident (may be not backed up with data, or still in process of getting detailed insight of a problem etc.). Whereas when I'm confident I tend to speak slowly or more assertively that made people to follow easily. I welcomed that feedback so from then on when I'm not confident in a topic I became more assertive to let pe"See full answer

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Mahima M. - "class Solution: def lengthOfLIS(self, nums: List[int]) -> int: temp = [nums[0]] for num in nums: if temp[-1]< num: temp.append(num) else: index = bisect_left(temp,num) temp[index] = num return len(temp) "See full answer

Kofi N. - "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"See full answer

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

Jeremy D. - "Any cycle would cause the prerequisite to be greater than the course. This passes all the tests: function canFinish(_numCourses, prerequisites) { for (const [a, b] of prerequisites) { if (b > a) return false } return true } `"See full answer

Devesh K. - "this solution here is much faster than the exponent reference soln. It is also far more concise and easy to understand def moveZerosToEnd(arr: List[int]) -> List[int]: left = 0 for right in range(len(arr)): if arr[right] == 0: pass else: if left != right: temp = arr[left] arr[left] = arr[right] arr[right] = temp left += 1 return arr `"See full answer

Anonymous Hare - "Sharing the approach for functional requirements we tool to solve this question. Functional Requirements This is only for the Registered users What is a "For You" page ? Home page where you get suggestions based on people you follow. Interactions like/share/comments (done by user) Interests (shared by the user during registration or onboarding) sports choices/ region choices/ Video sharing platform. So how many videos should we s"See full answer

Divya R. - "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"See full answer

Jerry O. - "\# Definition for a binary tree node. class TreeNode: def init(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def maxPathSum(self, root: TreeNode) -> int: self.max_sum = float('-inf')"See full answer

Prajwal M. - "from typing import List def maxprofitgreedy(stock_prices: List[int]) -> int: l=0 # buying r=1 # selling max_profit=0 while rstock_prices[l]: profit=stockprices[r]-stockprices[l] maxprofit=max(maxprofit,profit) else: l=r r+=1 return max_profit debug your code below print(maxprofitgreedy([7, 1, 5, 3, 6, 4])) `"See full answer

onering2ruleall - "My leadership style is flexible and adaptive, it varies depending on the team members and the needs of the company. My leadership goal is to empower the team and inspire and grow leaders. In order to achieve that, I combine transformational, democratic and coaching leadership styles. Usually when we are facing a new type of challenge, or at the early stage of a project, I like to adapt the transformational leadership which allows me to listen to all the suggestions from the team members and sta"See full answer

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

Tiago R. - "function findPrimes(n) { if (n < 2) return []; const primes = []; for (let i=2; i <= n; i++) { const half = Math.floor(i/2); let isPrime = true; for (let prime of primes) { if (i % prime === 0) { isPrime = false; break; } } if (isPrime) { primes.push(i); } } return primes; } `"See full answer